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Tema [17]
3 years ago
5

A rectification column is fed 100 kg mol/h of a mixture of 50 mol % benzene and 50 mol % toluene at 101.32 kPa abs pressure. The

feed is liquid at the boiling point. The distillate is to contain 90 mol % benzene and the bottoms 10 mol % benzene. The reflux ratio is 4.52:1. Calculate the kg mol/h distillate, kg mol/h bottoms, and the number of theoretical trays needed using the McCabe–Thiele method.

Engineering
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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3Px=y−y2p2<br><br>first order higher dgree​
dlinn [17]

Answer:

the order higher is 3p79g5t88=yv5379

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Which federal agency issues the model food code?
umka2103 [35]

The FDA! (food and drug administration)

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Read 2 more answers
Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°
cricket20 [7]

Answer:

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

m\cdot (h_{in} - h_{out}) = 0

h_{in} = h_{out}

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

P = 1500\,kPa

T = 198.29\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 6.3068\,\frac{kJ}{kg\cdot K}

x = 0.967

State 2 - Outlet (Superheated Vapor)

P = 200\,kPa

T = 130\,^{\textdegree}C

h = 2726.9\,\frac{kJ}{kg}

s = 7.1776\,\frac{kJ}{kg\cdot K}

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}

\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}

6 0
3 years ago
determine the position d of the 6- kn load so that the average normal stress in each rod is the same.
Zinaida [17]

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$\frac{3}{2} N_2 * x=N_2 *(L-x)

solving the above equation we have

$x=\frac{2}{5} L

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

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2 years ago
Only put coolant into your radiator when the engine is _____.
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Only put ciilant into ur radiator when the engine is cool (D)

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